Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 84  Maths Textbook Solution.

$\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|+C$

Hint:

You must know about integration of $\int \sqrt{a^{2}+x^{2}}$

Given:

$\int \sqrt{a^{2}+x^{2}} d x$

Solution:

$I=\int 1_{I I} \cdot \sqrt{a_{1}^{2}+x^{2}} d x$

$=\sqrt{a^{2}+x^{2}} \int 1 d x-\int\left(\frac{d}{d x}\left(\sqrt{a^{2}+x^{2}}\right) \int 1 d x\right) d x$

$=\sqrt{a^{2}+x^{2}} \cdot x-\int \frac{1 \times 2 x}{2 \sqrt{a^{2}+x^{2}}} \cdot x d x$

$=\sqrt{a^{2}+x^{2}} \cdot x-\int\left(\frac{x^{2}+a^{2}-a^{2}}{\sqrt{a^{2}+x^{2}}}\right) d x$

$=x \sqrt{a^{2}+x^{2}}-\int \sqrt{a^{2}+x^{2}} d x+a^{2} \int \frac{1}{\sqrt{a^{2}+x^{2}}} d x$

$=x \sqrt{a^{2}+x^{2}}-I+a^{2} \int \frac{1}{\sqrt{a^{2}+x^{2}}} d x$

$\therefore 2 I=x \sqrt{a^{2}+x^{2}}+a^{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|$

$\Rightarrow I=\frac{x}{2} \sqrt{a^{2}+x^{2}}+\frac{a^{2}}{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|+C$

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